Number Theory: Lecture 4

In which we study the structure of the multiplicative group (\mathbb{Z}/p\mathbb{Z})^{\times}, and start studying (\mathbb{Z}/p^2\mathbb{Z})^{\times}.

  • Theorem 15: Let p be prime.  Then the multiplicative group (\mathbb{Z}/p\mathbb{Z})^{\times} is cyclic.  We proved this by showing that there are exactly \phi(d) elements of order d, for each divisor d of p-1.
  • Definition of a primitive root modulo p.
  • Lemma 16: Let p be prime.  Then there is a primitive root modulo p, say g, such that g^{p-1} = 1 + bp where (b,p) = 1.  We showed that if a is a primitive root modulo p, then either a or a+p has the required additional property.
  • Lemma 17: Let p > 2 be prime and let j be a natural number.  Then there is a primitive root modulo p, say g, such that g^{p^{j-1}(p-1)} \not\equiv 1 \pmod{p^{j+1}}.  We proved this by induction, using similar ideas to those used in Lemma 16.

Understanding today’s lecture

Pick a prime p.  Can you find a generator (primitive root) modulo p?  How many can you find?  How little work can you get away with doing?  Can you write each element of the multiplicative group (\mathbb{Z}/p\mathbb{Z})^{\times} as an explicit power of your primitive root?

Can you find any primes that have many primitive roots, or any primes that have very few primitive roots?

Pick a prime p.  Can you find a generator (primitive root) modulo p^2?  Modulo p^3?

Can you find an example to show that in Lemma 16 we cannot always use our first choice of primitive root a, we really do sometimes have to use a+p instead?

Further reading

Davenport (The Higher Arithmetic) presents a slightly different proof that (\mathbb{Z}/p\mathbb{Z})^{\times} is cyclic; you might like to read that for another perspective.

Preparation for Lecture 5

Next time we’ll think about (\mathbb{Z}/p^j \mathbb{Z})^{\times}.  What structure will they turn out to have?

Advertisements

3 Responses to “Number Theory: Lecture 4”

  1. Tadek Krassowski Says:

    If g is a primitive root mod p but not mod p^2 we use g+p instead; but g and g+p are the same thing mod p. So I started wondering whether there always exists a primitive root mod p^2 smaller than p. I did not come up with anything and if it is not the case it will probably be pretty hard to find a counterexample since the chance of g not being a primitive root mod p^2 is 1/p. However, I have found a website (http://primes.utm.edu/curios/page.php/40487.html) according to which there are not many p^2 such that the smallest primitive root mod p is not a primitive root mod p^2 and, more interestingly, that it is not known whether there are infinitely many of those.

  2. Number Theory: Lecture 5 | Theorem of the week Says:

    […] Expositions of interesting mathematical results « Number Theory: Lecture 4 […]

  3. Tadek Krassowski Says:

    We proved in the supervision today that if g is a primitive root mod p, then either g or g^(-1) is a primitive root mod p^2. The proof is not difficult and it solves my problem from the post above.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s


%d bloggers like this: